Question

2 points It is known that the population standard deviation of students' debts is 11800 NIS. What is the minimum sample size should be selected so that the estimate at 95% confidence is within 800 NIS of the population mean? Your answer Given the sample: 15 18 30 40 40 44 50 50 Find the third quartile. * 2 points Your answer

Answer 1

Solution(1)
Given in the question
Population standard deviation($\sigma$) = 11800
The margin of error (E)= 800
Confidence level = 0.95
level of significance = 0.05, alpha/2 = 0.025
From Z table we found Zalpha/2 = 1.96
So sample size n can be calculated as
Sample size n = (Zalpha/2 * $\sigma$ /E)^2 = (1.96*11800/800)^2 = 835.7881 or 836
The minimum sample size should be 836 so that the estimate that 95% confidence is within 800 NIS.
Solution(2)
Quartile 3 can be calculated as
Q3 = 3*(n+1)/4
Here n = 3*(8+1)/4 = 6.75th value i.e. from the data set 6th value is 44 and 7th value is 50 so 6.75th value is 44+0.75*(50-44) = 48.5
So Q3 = 48.5